Solved on Dec 27, 2023

Differentiate e3x(sinx+2cosx)e^{3x}(\sin x + 2\cos x) with respect to xx.

STEP 1

Assumptions
1. We are differentiating the function f(x)=e3x(sinx+2cosx)f(x) = e^{3x}(\sin x + 2\cos x) with respect to xx.
2. We will apply the product rule for differentiation, which states that if u(x)u(x) and v(x)v(x) are differentiable functions of xx, then the derivative of u(x)v(x)u(x)v(x) is given by u(x)v(x)+u(x)v(x)u'(x)v(x) + u(x)v'(x).
3. We will also apply the chain rule for differentiation, which states that if y=g(u)y = g(u) and u=f(x)u = f(x) are differentiable functions, then dydx=dydududx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}.
4. We will use the basic derivatives that ddxeax=aeax\frac{d}{dx}e^{ax} = ae^{ax}, ddxsinx=cosx\frac{d}{dx}\sin x = \cos x, and ddxcosx=sinx\frac{d}{dx}\cos x = -\sin x.

STEP 2

Identify the two functions that are being multiplied in the given function.
Let u(x)=e3xu(x) = e^{3x} and v(x)=sinx+2cosxv(x) = \sin x + 2\cos x.

STEP 3

Differentiate u(x)u(x) with respect to xx using the chain rule.
u(x)=ddxe3x=3e3xu'(x) = \frac{d}{dx}e^{3x} = 3e^{3x}

STEP 4

Differentiate v(x)v(x) with respect to xx using the sum rule and the basic derivatives of sinx\sin x and cosx\cos x.
v(x)=ddx(sinx+2cosx)=cosx2sinxv'(x) = \frac{d}{dx}(\sin x + 2\cos x) = \cos x - 2\sin x

STEP 5

Apply the product rule to differentiate f(x)=u(x)v(x)f(x) = u(x)v(x).
f(x)=u(x)v(x)+u(x)v(x)f'(x) = u'(x)v(x) + u(x)v'(x)

STEP 6

Substitute u(x)u(x), v(x)v(x), u(x)u'(x), and v(x)v'(x) into the product rule formula.
f(x)=(3e3x)(sinx+2cosx)+(e3x)(cosx2sinx)f'(x) = (3e^{3x})(\sin x + 2\cos x) + (e^{3x})(\cos x - 2\sin x)

STEP 7

Distribute 3e3x3e^{3x} through the first set of parentheses and e3xe^{3x} through the second set of parentheses.
f(x)=3e3xsinx+6e3xcosx+e3xcosx2e3xsinxf'(x) = 3e^{3x}\sin x + 6e^{3x}\cos x + e^{3x}\cos x - 2e^{3x}\sin x

STEP 8

Combine like terms.
f(x)=(3e3xsinx2e3xsinx)+(6e3xcosx+e3xcosx)f'(x) = (3e^{3x}\sin x - 2e^{3x}\sin x) + (6e^{3x}\cos x + e^{3x}\cos x)

STEP 9

Simplify the expression by combining the coefficients of like terms.
f(x)=e3xsinx+7e3xcosxf'(x) = e^{3x}\sin x + 7e^{3x}\cos x

STEP 10

Write the final simplified expression for the derivative of the given function.
ddxe3x(sinx+2cosx)=e3xsinx+7e3xcosx\frac{d}{dx}e^{3x}(\sin x + 2\cos x) = e^{3x}\sin x + 7e^{3x}\cos x
The derivative with respect to xx of the function e3x(sinx+2cosx)e^{3x}(\sin x + 2\cos x) is e3xsinx+7e3xcosxe^{3x}\sin x + 7e^{3x}\cos x.

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